A 0.02kg bullet is fired at a 3.5kg block which is initially at resr on a table. The bullet embeds in the block. Friction converts the 24.02J of kinetic energy to heat and brings the block and bullet system to stop.

Question

A 0.02kg bullet is fired at a 3.5kg block which is initially at resr on a table. The bullet embeds in the block. Friction converts the 24.02J of kinetic energy to heat and brings the block and bullet system to stop.
How fast was the bullet and block system sliding immediately after the collision?

ChrismB · Answer

i know its answer, which is 3.74, but i have no idea how does it can be 3.74

Chanz · Answer

KE = 1/2 mv^2 Solve for v I get 3.69